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Re^2: Question on simple arithmetic using perl

 on Sep 28, 2011 at 21:43 UTC ( #928434=note: print w/replies, xml ) Need Help??

Thank you Monk!! So does this mean its fine to use such arithmetic and not worry about whether perl can handle it correctly?
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Re^3: Question on simple arithmetic using perl
by Marshall (Abbot) on Sep 28, 2011 at 23:08 UTC
No. A float can represent a wider range of values, but a lesser number of significant digits than an integer.

I'm not quite sure what the question is:

```#!/usr/bin/perl -w
use strict;

my \$all_bits_are_on = -1;
printf ("all_bits => %X\n", \$all_bits_are_one);

my \$x = \$all_bits_are_on + 1;
print "Adding -1 to +1 is: \$x\n";

__END__
all_bits => FFFFFFFF
Adding -1 to +1 is: 0

Note that perl is actually using long and double. On 32bit double has more significant digits than long.

Yes, if you mean that a 64 bit floating point number has more significant bits in the mantissa than a 32 bit integer does, I agree. I leave looking up the definition of ANSI C long and double to the interested readers.

Update: I looked up what Microsoft did for storage in C. "int" and "long" are both 32 bits (other folks will assign 64 bits to a "long"). A MS "float" is 32 bits and MS "double" is 64 bits with 52 bit mantissa.

ANSI C Summary says: that an int will have at least 4 significant decimal digits and a long will have at least 9 significant decimal digits (see link for exact ANSI spec'd ranges). An int can be the same size as a long (bigger than minimum is ok).

A float has to have 6 decimal digits of precision and a double has a minimum of 10 decimal digits of precision. So a double is guaranteed to have at least one more decimal digit of precision than a long (10 vs 9).

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